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I just have a quick question!

We got a question asking us which of the following algebraic structures were fields, I got every one of them except the last one. I don't even know how to begin how to attack the question.

Is real numbers modulo 2$\pi$ a field?

user70871
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  • Multiplication is not even well-defined. We can easily have $x,x'$ differing by an integer multiple of $2\pi$, same for $y,y'$, but where $xy$ and $x'y'$ do not differ by an integer multiple of $2\pi$. – André Nicolas Apr 10 '13 at 19:00
  • @AndréNicolas Everything is a real multiple of $2\pi$. Multiplication sends $0 \cdot 0$ to $0$. – Cocopuffs Apr 10 '13 at 19:01
  • @Cocopuffs: I managed to sneak in under the $5$ minute time limit and change multiple to what I meant, integer multiple. – André Nicolas Apr 10 '13 at 19:04
  • @AndréNicolas Yes, but why do you want integer multiples? – Cocopuffs Apr 10 '13 at 19:04
  • Possible duplicate: http://math.stackexchange.com/q/355001/65389 – A.P. Apr 10 '13 at 19:05
  • Because of the natural additive group structure. – André Nicolas Apr 10 '13 at 19:06
  • I don't know if this would help, but my teacher gave us this blurb at the end saying '(ie: the set of equivalence classes of real numbers under the equivalence relation given by $x-y \in 2 \pi \mathbb{Z}$)' – user70871 Apr 10 '13 at 19:06
  • Then you mean the real numbers modulo $2\pi \mathbb{Z}$. That might be worth adding in the question. – Cocopuffs Apr 10 '13 at 19:07
  • Then the quotient object you are referring to is isomorphic to $\Bbb R/\Bbb Z$, which is the subject of the question I linked. – A.P. Apr 10 '13 at 19:08
  • You might want to increase your odds of getting help by accepting an answer for each question you've asked, if and when one has been helpful. You can accept one answer per question. To accept an answer, you can click on the $\checkmark$ to the left of the answer you'd like to accept. You also receive two reputation points for each answer you accept. If you have questions you'd like to ask about an answer, you can do so by commenting below the answer. – amWhy Apr 12 '13 at 03:20
  • IMHO this question is not a duplicate. It is clear from the OP's comments to answers of this question that the level of the answer of the linked question is not accessible to the OP; so this question generated new explanations that were more geared to the OP's prior knowledge. This enriches the site as well as allowing it to function in answering the OP's question. – Ben Blum-Smith Apr 12 '13 at 20:29

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If you want the quotient to be a ring, then you must take the quotient by an ideal. The ideal generated by $2\pi$ is all of $\Bbb R$, so this quotient is the zero ring, which is not considered to be a field.

This is nothing special with $2\pi$: every nonzero number will make the quotient zero. If you try to mod out by zero, you will of course get just $\Bbb R$ back, and so that would be a field!

If you have convinced yourself the thing you are modding out is not all of $\Bbb R$, then unfortunately you are thinking of something which isn't an ideal, and so the quotient by this object will not even be a ring (and of course could not be a field, then.)

rschwieb
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  • Unfortunately, I'm not familiar with rings :( Is there possibly another way to explain this? – user70871 Apr 10 '13 at 19:04
  • A ring is what you get when you think of a field, but you decide to forget about multiplicative inverses and commutative multiplication. It is not a good idea for whoever assigned this to you to give a question about a quotient but not explain what the quotient in the problem meant... – rschwieb Apr 10 '13 at 19:07
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It's not a field. In fact, if $\mathbb{R}$ mod $2\pi$ is referring to the group of radian angle measures under addition, then it doesn't even have a good multiplication operation. It has well-defined addition, but multiplication is a problem because the remainder you get when subtracting the closest integer multiple of $2\pi$ depends on what representative you pick mod $2\pi$. For example, $$1\times 1 = 1$$

However,

$$(1+2\pi)\times(1+2\pi)=1+4\pi + 4\pi^2$$

which does not differ by an integer multiple of $2\pi$ from $1$. In the example of angle measures, $1$ and $1+2\pi$ "should" represent the same angle, but there is no good way to "multiply angles" so that $1$ and $1+2\pi$ will be guaranteed to get the same answer.

EDIT: a propos of your comment to rschweib, "ring" is the same definition as "field" except elements aren't required to have multiplicative inverses so there's not necessarily a good division operation. For example, the integers form a ring because you can add, subtract and multiply, but not necessarily divide. To be a field, you also need to be able to divide. When we say, "it's not even a ring!" we mean not only do we not have division, we don't even have multiplication.

Ben Blum-Smith
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For $K$ to be a field, you need to first state: "What is the set, what is the addition operator, and what is the multiplication operator?" Once you have these understood, you need to pass the following two tests:

  1. Is $K$ an abelian group under +?
  2. Is $K - \{0\}$ an abelian group under $\times$?

So in your specific example, the set is infinite: it's $\mathbb{R} \bmod 2\pi$, which is essentially the interval $[0, 2\pi)$. The addition and multiplication operators are simply the same ones from $\mathbb{R}$ but modulo $2\pi$.

So now apply the two tests above. Is $[0,2\pi)$ an abelian group under addition mod $2\pi$? You have to go through the requirements for abelian groups (well-defined operation, closure, associativity, existence of an identity, and inverses for all elements). You should be able to convince yourself that the answer is "yes."

Now do this again for the second requirement: is $(0,2\pi)$ an abelian group under multiplication mod $2\pi$? (Notice I changed the left bracket to a paren to indicate the exclusion of 0 here.) As others have already pointed out, the operation is not well-defined: by saying that our set is "equivalent" to the interval $(0,2\pi)$ we are grouping all multiples of $2\pi$ together into one equivalence class called $0$ (because after all, $0=2\pi =4\pi =6\pi= \cdots = -2\pi = -4\pi=\cdots$ when considered mod $2\pi$). In order for multiplication to be "well defined" it must be true that you can choose any representation of a set member and multiply it by any other set member and always get the same result. But, as Ben Blum-Smith already pointed out, this fails in our setting. So multiplication is not well-defined and we don't have an abelian group. (Once you find the operation is not well-defined, you can't even sensibly ask about the other group properties, so you stop here.)

Of course this is all very long-winded. If you are taking a test, you would immediately suspect this thing isn't a field. You don't have to find all the reasons why, just find one reason. And with some experience, you would quickly "sense" that something isn't going to work with multiplication and you would zero in on this "not well-defined" problem.

Fixee
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The circle (or torus) quotient group $\rm\ R/G = \Bbb R/2\pi\Bbb Z\ $ does not form a ring, because, generally

Theorem $\ $ Coset multiplication $\rm\:(r\! +\! G)(s\! +\! G) = rs\! +\! G\:$ is well-defined $\rm\!\iff\! G\,$ is an ideal of $\rm\,R$

Thus the additive quotient group $\rm\:R/G\:$ inherits the multiplication of $\rm\:R\:$ iff $\rm\:G,\:$ in addition to being an additive subgroup of $\rm\,R,\:$ is, $ $ further, $ $ closed under multiplication by all elements of $\rm\,R.$

In your example, it is easy to construct explicit examples illustrating that such coset multiplication is not well-defined.

Math Gems
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